Instantaneous acceleration *acceleration at a specific instant in time*, is obtained by the same process as discussed for instantaneous velocity in Time, Velocity, and Speed—that is, by considering an infinitesimally small interval of time. How do we find instantaneous acceleration using only algebra? The answer is that we choose an average acceleration that is representative of the motion. Figure 6 shows graphs of instantaneous acceleration versus time for two very different motions. In Figure 6(a), the acceleration varies slightly and the average over the entire interval is nearly the same as the instantaneous acceleration at any time. In this case, we should treat this motion as if it had a constant acceleration equal to the average (in this case about

The next several examples consider the motion of the subway train shown in Figure 7. In (a) the shuttle moves to the right, and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems.

### Example 2: **Calculating Displacement: A Subway Train**

What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure 7?

*Strategy*

A drawing with a coordinate system is already provided, so we don’t need to make a sketch, but we should analyze it to make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation

*Solution*

1. Identify the knowns. In the figure we see that

2. Solve for displacement in part (a).

3. Solve for displacement in part (b).

*Discussion*

The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus has a negative sign.

### Example 3: **Comparing Distance Traveled with Displacement: A Subway Train**

What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in Figure 7?

*Strategy*

To answer this question, think about the definitions of distance and distance traveled, and how they are related to displacement. Distance between two positions is defined to be the magnitude of displacement, which was found in Example 2. Distance traveled is the total length of the path traveled between the two positions. (See Displacement.) In the case of the subway train shown in Figure 7, the distance traveled is the same as the distance between the initial and final positions of the train.

*Solution*

1. The displacement for part (a) was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km, and the distance traveled was 2.00 km.

2. The displacement for part (b) was

*Discussion*

Distance is a scalar. It has magnitude but no sign to indicate direction.

### Example 4: **Calculating Acceleration: A Subway Train Speeding Up**

Suppose the train in Figure 7(a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?

*Strategy*

It is worth it at this point to make a simple sketch:

This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration.

*Solution*

1. Identify the knowns.

2. Calculate

3. Plug in known values and solve for the unknown,

4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See Physical Quantities and Units for more guidance.)

*Discussion*

The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the *change* in velocity, as is always the case.

### Example 5: **Calculate Acceleration: A Subway Train Slowing Down**

Now suppose that at the end of its trip, the train in Figure 7(a) slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration while stopping?

*Strategy*

In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration.

*Solution*

1. Identify the knowns.

2. Solve for the change in velocity,

3. Plug in the knowns,

4. Convert the units to meters and seconds.

*Discussion*

The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion. Again, acceleration is in the same direction as the *change* in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction opposite to the velocity.

The graphs of position, velocity, and acceleration vs. time for the trains in Example 4 and Example 5 are displayed in Figure 10. (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.)

### Example 6: **Calculating Average Velocity: The Subway Train**

What is the average velocity of the train in part b of Example 2, and shown again below, if it takes 5.00 min to make its trip?

*Strategy*

Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement.

*Solution*

1. Identify the knowns.

2. Determine displacement,

3. Solve for average velocity.

4. Convert units.

*Discussion*

The negative velocity indicates motion to the left.

### Example 7: **Calculating Deceleration: The Subway Train**

Finally, suppose the train in Figure 11 slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration?

*Strategy*

Once again, let’s draw a sketch:

As before, we must find the change in velocity and the change in time to calculate average acceleration.

*Solution*

1. Identify the knowns.

2. Calculate

3. Solve for

4. Convert units.

*Discussion*

The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to the left) in this problem and a positive acceleration opposes the motion (and so it is to the right). Again, acceleration is in the same direction as the *change* in velocity, which is positive here. As in Example 5, this acceleration can be called a deceleration since it is in the direction opposite to the velocity.